Question

Elemental analysis results obtained for cortisone, an anti-inflammatory agent, are 69.98% C, 7.83 H and 22.19% O. What is the empirical formula of cortisone?

• A. $C_4{H}_{6}O$
• B. $C_{18}{H}_{22}{O}_3$
• C. $C_{20}{H}_{25}{O}_4$
• D. $C_{12}{H}_{28}{O}_5$

Answer 1

Answer: (A)

$C_4{H}_{6}O$

100 g of the compound will contain 69.98 g $C$, 7.83 g $H$ and 22.19 g $O$ respectively.

The atomic masses of $C$, $H$ and $O$ are 12 g/mol, 1 g/mol and 16 g/mol respectively.

Number of moles of $C=\frac{69.98g}{12g/mol}=5.832mol$

Number of moles of $H=\frac{7.83g}{1g/mol}=7.83mol$

Number of moles of $O=\frac{22.19g}{16g/mol}=1.39mol$

The ratio of the number of moles $C:H:O=5.832:7.83:1.39=4.2:5.6:1$

This is approximately equal to $4:6:1$

Hence, the empirical formula is $C_4{H}_{6}O$